Q.Calculate the ratio of the frequencies of the radiation emitted due to transition of the electron in a hydrogen atom from its

Why the Bohr Model Gives Those Energy Levels

The Bohr model is a beautiful piece of physics because it takes a simple, almost desperate idea — "electrons only exist in certain orbits" — and derives the entire hydrogen spectrum from it. The key is that Bohr didn't just assume the energy levels; he forced them to be consistent with classical physics in one specific way, then broke with it in another.

The Two Non-Negotiable Pieces

First, the electron moves in a circle around the proton. That's pure classical mechanics: the Coulomb attraction provides the centripetal force.

This gives you a relation between speed $v$ and radius $r$:

Second, the total energy of the electron is the sum of its kinetic and potential energies. Potential energy for a Coulomb force is $-\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}$ (negative because the force is attractive, and we set zero at infinity).

Substitute $v^2$ from above:

So far, nothing is quantised. Any radius $r$ gives a valid classical orbit, and the energy just follows from that radius. The problem is that a classical electron in a curved path radiates energy and spirals into the nucleus — atoms should collapse. Bohr needed a rule to pick out stable orbits.

The Quantisation Condition

Bohr's revolutionary step was to postulate that the angular momentum of the electron is quantised in units of $\hbar = h/2\pi$:

Why this particular rule? Bohr later said it was the simplest way to get the right answer. But there's a deeper physical motivation: if you think of the electron as a wave (de Broglie's idea, which came a decade later), the condition that a standing wave fits exactly around the circumference $2\pi r = n \lambda$ gives $m v r = n \hbar$ directly. So the quantisation condition is really a wave condition imposed on a particle picture.

Deriving the Allowed Radii and Energies

From $m v r = n \hbar$, we get $v = n\hbar / (mr)$. Substitute this into the centripetal force equation:

Solve for $r$:

The constant in front is the Bohr radius $a_0 \approx 0.529 \, \text{Å}$. So the radii are $r_n = a_0 n^2$.

Now plug $r_n$ back into the energy expression $E = -\frac12 \frac{e^2}{4\pi\varepsilon_0 r}$:

Substitute $a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m e^2}$: